Everyone seems a bit confused about what you are trying to do. You are trying to prove that lemma in order to prove Schroder-Cantor-Bernstein (SCB). I will demonstrate that there's no real way to do that, i.e. there is an effectively trivial reduction making SCB and your lemma equivalent. Thus any proof of this is effectively a proof of SCB, which is somewhat technical and challenging. E.g. see here for a proof. Since there is no (known) easy (by undergrad standards) proof of SCB, there is no such proof of your statement either. Thus you are effectively asking "how do I prove SCB?", for which you can see above or, since we like to keep things in-site, here.
So why are these problems the same? For the sake of brevity, let's call your lemma L. Also for notation, write $C\cong D$ for there exists a bijection between $C$ and $D$. It is easy to verify $\cong$ is an equivalence relation.
Suppose we know L and we want to show SCB. Let $A, B$ sets with injections $f: A \to B$ and $g : B \to A$. Then clearly $B \cong g(B)$ (under the bijection of $g$ with restricted range). Then we have a bijection $h: g(B) \to B$, in particular a surjection. Also $f$ has a left inverse $k: B \to A$ (i.e. $k \circ f = id$) (for proof see here). Then $k$ is a surjection, as can be easily checked. Thus $k \circ h : g(B) \to A$, as a composition of surjections is a surjection. Thus we have a surjection $g(B) \to A$ so by your lemma, L, we have $g(B) \cong A$. Thus $A \cong B$. So we shown SCB.
Now suppose we know SCB. Let $A \subseteq B$ and suppose $f: A \to B$ is a surjection. Since surjections have right inverses there is some $g: B \to A$ s.t. $f \circ g = id$. Thus $g$ is an injection. So we have an injection $B \to A$. But the inclusion map $i: A \to B$ just given by $i(x) = x$ is clearly and injection. Thus SCB applies and $A \cong B$. Thus we have shown your lemma, L.
